Question

By using the method of completing the square, show that the equation $4x^2+3x+5=0$ has no real roots.

Answer 1

We have
$4x^2+3x+5=0$

$\Rightarrow x^2+\frac{3}{4}x+\frac{5}{4}=0\Rightarrow x^2+2\left(\frac{3}{8}x\right)x=-\frac{5}{4}\Rightarrow x^2+2\left(\frac{3}{8}x\right)+{\left(\frac{3}{8}\right)}^2={\left(\frac{3}{8}\right)}^2-\frac{5}{4}\Rightarrow {(x+\frac{3}{8})}^2=-\frac{71}{64}$
Clearly, RHS is negative. But ${(x+\frac{3}{8})}^2$ cannot be negative for any real value of $x$. Hence, the given equation has no real roots.